Answer:
169.74 N
Explanation:
Given,
Mass of the girl = 30 Kg
angle of the rope with vertical, θ = 30°
equating the vertical component of the tension
vertical component of the tension is equal to the weight of the girl.
T cos θ = m g
T cos 30° = 30 x 9.8
T = 339.48 N
Tension on the two ropes is equal to 339.48 N
Tension in each of the rope = T/2
= 339.48/2 = 169.74 N
Hence, the tension in each of the rope is equal to 169.74 N
Answer:
The value of the centripetal forces are same.
Explanation:
Given:
The masses of the cars are same. The radii of the banked paths are same. The weight of an object on the moon is about one sixth of its weight on earth.
The expression for centripetal force is given by,

where,
is the mass of the object,
is the velocity of the object and
is the radius of the path.
The value of the centripetal force depends on the mass of the object, not on its weight.
As both on moon and earth the velocity of the cars and the radii of the paths are same, so the centripetal forces are the same.
Answer: 25.38 m/s
Explanation:
We have a straight line where the car travels a total distance
, which is divided into two segments
:
(1)
Where 
On the other hand, we know speed is defined as:
(2)
Where
is the time, which can be isolated from (2):
(3)
Now, for the first segment
the car has a speed
, using equation (3):
(4)
(5)
(6) This is the time it takes to travel the first segment
For the second segment
the car has a speed
, hence:
(7)
(8)
(9) This is the time it takes to travel the secons segment
Having these values we can calculate the car's average speed
:
(10)
(11)
Finally:
Answer:
is the initial velocity of tossing the apple.
the apple should be tossed after 
Explanation:
Given:
- velocity of arrow in projectile,

- angle of projectile from the horizontal,

- distance of the point of tossing up of an apple,

<u>Now the horizontal component of velocity:</u>



<u>The vertical component of the velocity:</u>



<u>Time taken by the projectile to travel the distance of 30 m:</u>



<u>Vertical position of the projectile at this time:</u>



<u>Now this height should be the maximum height of the tossed apple where its velocity becomes zero.</u>


is the initial velocity of tossing the apple.
<u>Time taken to reach this height:</u>



<u>We observe that </u>
<u> hence the time after the launch of the projectile after which the apple should be tossed is:</u>



Answer:
37.8 m
Explanation:
At point 0, the ball is at height y₀.
At point 1, the ball is at height 30 m.
At point 2, the ball is at height 0 m.
Given:
y₁ = 30 m
y₂ = 0 m
v₀ = 0 m/s
a = -10 m/s²
t₂ − t₁ = 1.5 s
Find: y₀
Use constant acceleration equation.
y = y₀ + v₀ t + ½ at²
Evaluate at point 1.
y₁ = y₀ + v₀ t₁ + ½ at₁²
30 m = y₀ + (0 m/s) t₁ + ½ (-10 m/s²) t₁²
30 = y₀ − 5t₁²
Evaluate at point 2.
y₂ = y₀ + v₀ t₂ + ½ at₂²
0 m = y₀ + (0 m/s) t₂ + ½ (-10 m/s²) t₂²
0 = y₀ − 5t₂²
y₀ = 5t₂²
Substitute:
y₀ = 5 (1.5 + t₁)²
y₀ = 5 (2.25 + 3t₁ + t₁²)
y₀ = 11.25 + 15t₁ + 5t₁²
30 = 11.25 + 15t₁ + 5t₁² − 5t₁²
30 = 11.25 + 15t₁
t₁ = 1.25
30 = y₀ − 5t₁²
30 = y₀ − 5(1.25)²
y₀ ≈ 37.8